Question

A body falls freely from the top of a tower. It covers 36% of the total height in the last second before striking the ground level , the height of the tower is
a)50m b)75m c)100m d)125m

Answer 1

Given:-

→A body falls freely from the top of a

tower.

→It covers 36% of the total height in the

last second before striking ground.

To find:-

→Height of the tower

Solution:-

•Initial velocity of the body will be zero as it was initially at rest

•We will take g = +10m/s² to get an appropriate answer.

By using the 2nd equation of motion,we get:-

=>h = ut+1/2gt²

=>h = 1/2×10×t²

=>h = 5t². ----(1)

Now, the body covers 36% of the total height in the last second, thus it covers 64% of the total height in (t-1) seconds.

=>64/100×h = 0(t)+1/2×10×(t-1)²

=>0.64×5t² = 5(t-1)²

=>0.64 = 5(t-1)²/5t²

=>0.64 = (t-1)²/t²

=>0.64t² = (t-1)²

=>√0.64t² = √(t-1)²

=>0.8t = t-1

=>0.8t-t = -1

=>-0.2t = -1

=>t = -1/-0.2

=>t = 5s

Thus,we got t = 5s

Now,by putting the value of t in eq.1,we get:-

=>h = 5(5)²

=>h = 5(25)

=>h = 125m

Thus,height of the tower is 125m (Option.d)

Answer 2

Let the height of the tower be h. Let the mass of the ball be m.

Let $\sf{g=10\ m\,s^{-2}.}$

The body is dropped from the top of the tower, so it's initial velocity is zero.

By third equation of motion, the velocity of the body when it reaches ground level is given by,

$\sf{\longrightarrow v^2=0^2+2gh}$

$\sf{\longrightarrow v^2=20h}$

$\sf{\longrightarrow v=\sqrt{20h}}$

Before the last second of striking the ground level, the height of the body is 36% of the total height (height of the tower), since the body covers this much distance in this last second as per the question.

The body comes to this position after travelling 64% of the total height.

So by third equation of motion, the velocity of the body at this height is given by,

$\sf{\longrightarrow u^2=0^2+2g\cdot\dfrac{64h}{100}}$

$\sf{\longrightarrow u^2=12.8h}$

$\sf{\longrightarrow u=\sqrt{12.8h}}$

So the body travels 36% of the total height in the last second.

By first equation of motion,

$\sf{\longrightarrow v=u+gt}$

$\sf{\longrightarrow\sqrt{20h}=\sqrt{12.8h}+10}$

$\sf{\longrightarrow\sqrt h(\sqrt{20}-\sqrt{12.8})=10}$

$\sf{\longrightarrow0.89\sqrt h=10}$

$\sf{\longrightarrow\sqrt h=11.18}$

$\sf{\longrightarrow\underline{\underline{h=125\ m}}}$

Hence (d) is the answer.